Mathematics Paper 2 Questions and Answers - Momaliche Pre Mock Exams 2023

INSTRUCTIONS TO CANDIDATES

• This paper contains TWO sections: section I and section II
• Answer ALL the questions in Section I and only five questions from section II.
• Show all the steps in your calculations, giving your answers at each stage in the spaces provided below each question.
• Marks may be given for correct working even if the answer is wrong.
• Non-programmable silent electronic calculators and KNEC mathematical tables may be used except where stated otherwise.

SECTION 1 (50 MARKS)

Answer all questions in this section in the spaces provided.

1. A positive two digit number is such that the product of the digits is 24. When the digits are reversed, the number formed is greater than the original number by 18. Find the number.          (3mks)
2. Use tables of squares, square roots and reciprocals to evaluate                       (4mks)
        234     +        16       
   √0.02698     (0.18149)²
3. The height and radius of a cone are measured as 21 cm and 14.0 cm respectively. Taking π= 3.142, find the percentage error in the volume of the cone.                               (3mks)
4. Express the following in surd form and simplify by rationalizing the denominator without using a calculator and leave your answer in the form  a + b√c       (3mks)                  
   1 + Cos 30°
    1− Sin 60°
5. Solve for x in : Log₂(x + 7) – Log₂(x – 7) = 8                                                        (3mks)
6. A businessman obtained a loan of Ksh 450,000 from a bank to buy a Matatu that was valued at the same amount. The bank charges interest at 24% per annum compounded quarterly per year. Calculate the total amount of money the businessman paid to clear the loan in 4½ years to the nearest shilling.       (3mks)
7. In the diagram below, BT is a tangent to the circle at B. AXCT and BXD are straight lines. AX = 6cm, CT = 8cm, BX = 4.8cm and XD = 5cm.
   
   Find the length of BT.    (3Marks)
8. Find the possible values of x given that     is a singular matrix.        (3mks)
9. The cost C of operating an electronic business is partly constant and partly varies as the square of labour input L. If C=25,000 when L=20 and C=45,000 when L=30. Find C when L=8.    (3Mks)
10. The 2^nd, 4^thand 7^th terms of an A.P. are the first 3 consecutive terms of a G.P. Find the common ratio of the G.P if the common difference of the A.P. is 2.      (3mks)
11. P and Q are two points such that OP=i + 2j + 3k and OQ = 4i + 5j – 3k. M is a point that divides PQ externally in the ratio 3:2. Find the co-ordinates of M, given that O is the origin.                     (3mks)
12. A circle Centre C (5, 5) passes through points A (1, 3) and B (a, 9). Find the equation of the circle and hence the possible values of a.                              (4mks)                                                                          
13. Tap A can fill an empty tank in 3 hours, while tap B can fill the same tank in 2 hours. When the tank is full, tap C can empty the tank in 5 hours. Tap A and C are opened for 4 hours and then closed.
    1. Determine the fraction of the tank that is still empty.                          (1mks)
    2. Find how long it would take to fill the remaining fraction of the tank if all the three taps are opened.     (2mks)                                              
14. Determine the interquartile range for the following set of numbers.                         (2mks)
    4, 9, 5, 4, 7, 6, 2, 1, 6, 7, 8.
15. Solve the equation Sin(3x – 10)° =  0.4337 for 0°≤ Ɵ ≤180°                               (3mks)
16.  
    1. Expand and simplify (3x – y)⁴(2mks)
    2. Use the first three term of the expansion to approximate the value of (6 – 0.2)^4.  (2mks)

SECTION II (50MARKS) ANSWER ANY 5 QUESTIIONS ONLY

17. Mrs. Mutua earns a basic salary of K£ 12,000 p.a. and is housed by the employer at a nominal rent of Shs 1,200 per month. She is entitled to a personal relief of K£ 1,320 p.a. and a premium relief of 10% on her insurance premium of K£ 800 p.a. The table of tax rate is as below.

    Taxable income (K£ p.a.)	Rate (%)
    1 – 2100	10
    2101 – 4200	15
    4201 – 6300	20
    6301 – 8400	25
    Over     8400	30

    Calculate;
    1. Calculate the net tax per annum.                                (7mks)
    2. Other deductions includes W.C.P.S  Shs 600 per month, NHIF Shs. 500 per month. Calculate her net pay per month.   (3mks)
18. The Line AB = 5cm is a side of a triangle ABC in which angle ABC = 90⁰ and angle BAC = 60⁰.
    1. Construct triangle ABC (2mks)
    2. Construct the Locus P such that angle APB = angle ACB (2mks)
    3. Locate by construction points Q1 and Q2 which satisfy the conditions below:
       1. Q1 and Q2 lie on the same side of line AB and C(3mks
       2. Area of triangle AQ1B = Area of triangle AQ2B = ¾Area of triangle ABC
       3. Angle AQ1B = Angle AQ2B = 30⁰
          Measure the length of the line Q1Q2 (3mks)
    4. Calculate the area above the line Q1Q2 bounded by the locus of point P       (3mks)
19. The diagram below shows a square based pyramid V vertically above the middle of the base.  PQ = 10cm and VR = 13cm.  M is the midpoint of VR.
    
    Find to 2 decimal places
    1. 1. the length PR.   (2mks)
       2. The height of the pyramid.      (2mks)
    2.  
       1. the angle between VR and the base PQRS.    (2mks)
       2. The angle between MR and the base PQRS.       (2mks)
       3. The angle between the planes QVR and PQRS.                                (2mks)
20.  
    1. Complete the table below for y = sin 2x and y = sin ( 2x + 30) giving values to 2d.p(2mks)
       

       X	0	15	30	45	60	75	90	105	120	135	150	165	180
       Sin 2x	0				0.87				−0.87				0
       Sin ( 2x +30)	0.5				0.5				−1				0.5

    2. Draw the graphs of y=sin 2x and y = sin (2x + 30) on the axis.(4mks)
    3. Use the graph to solve  sin (2x + 30) − sin 2x = 0. (1mk)
    4. Determine the transformation which maps sin 2x onto sin (2x + 30).  (1mk)
    5. State the period and amplitude ofy = sin (2x + 30)     (2mks)
21. OABC is a parallelogram with verities 0(0,0), A(2,0)  B(3,2) and C(1,2). O,A,B,C   is the image of OABC under transformation matrix.
    1.  
       1. Find the coordinates of O¹A¹B¹C^1.    (2mks)
       2. On the grid provided, draw OABC and O¹A¹B¹C¹    (2mks)
    2.    
       1. Find O¹¹A¹¹B¹¹C¹¹, the image of O¹A¹B¹C¹ under transformation matrix (2mks)
       2. On the same grid draw O¹¹A¹¹B¹¹C ¹¹ (1mk)
    3. Find a single matrix that maps O¹¹A¹¹B¹¹C¹¹ onto OABC               (3mks)
22. The following table shows the distribution of marks obtained by 50 students in a test.
    

    Marks	45-49	50-54	55-59	60-64	65-69	70-74	75-79
    No. of Students	3	9	13	15	5	4	1

    By using an assumed mean of 62, calculate
    1. The mean.  (5mks)
    2. The variance. (3mks)
    3. The standard deviation.   (2mks)
23. A box contains 3 brown, 9 pink and 15 white cloth pegs.  The pegs are identical except for the colour.  
    1. Find the probability of picking.
       1. A brown peg.   (1mark)
       2. A pink or a white peg.  (2 marks)
    2. Two pegs are picked at random, one at a time without replacement.  Find the probability that:
       1. Atleast one brown peg is picked.   (4marks)
       2. both pegs are of the same colour.   (3marks)
24. A wholesaler stocks two types of rice: Refu and Tamu. The wholesale prices of 1 kg of Refu and 1 kg of Tamu are Ksh 80 and Ksh 140 respectively. The wholesaler also stocks blend A rice which is a mixture of Refu and Tamu rice mixed in the ratio 3 : 2.
    1.  
       1. A retailer bought 10 kg of blend A rice. To this blend, the retailer added some Tamu rice to prepare a new mixture blend X. The .ratio of Refu rice to Tamu rice in blend X was 1:2.
          Determine the amount of Tamu rice that was added.   (3marks)
       2. The retailer sold blend X rice making a profit of 20%. Determine the selling price of 1 kg of blend X.  (3 marks)
    2. The wholesaler prepared another mixture, blend B, by mixing x kg of blend A rice with y kg of Tamu rice. Blend B has a wholesale price of Ksh130 per kg.Determine the ratio x : y.         (4mks)                                     

MARKING SCHEME

5	Log2x+7 =Log28           M1(for correct Log28)         x-7 Dropping the logs x+7 = 8                                                        B1(for dropping logs) x-7 x + 7 = 8x – 56 63 = 7x X = 9                                                           A1
	Total 3 mks
6	A = 450,000(1 + 24/100 x ¼)18               M(for correct substitution) A= 1,284,453                                                       A1
	Total 2 mks
7	AX.CX = DX.XB 6XC = 5 × 4.8    M1 XC = 5 × 4.8 = 4 cm   M1             6 BT2= 8 × 18 BT = √144 = 12     A1
	Total 3mks
8	X2 + 8x – 48 = 0  M1 Product = -48 and sum = 8 Factors (12, -4) X2 + 12x – 4x – 48 = 0 Type equation here. X(x + 12) -4(x +12) = 0       M1(for solving quadratic equation) Either x + 12 = 0 or x – 4 = 0 X= −12   or  x = 4  A1
	Total 3mks
9	C = k + hL2 25,000= k + 400h 45,000= k + 900h −20,000 = −500h h= 40   M1 25,000= k + 400(40) k = 9,000  M1 C = 9,000 + 40(82) C= Ksh. 11,560           A1
	Total 3mks
10	2nd = a + d 4th = a + 3d 7th = a + 6d a + 3d = a + 6d    M1  a + d     a + 3d (a + 3d)( a + 3d)=( a + 6d)( a + d) a2 + 6ad + 9d2=a2 + 7ad + 6d2 3d2= ad     hence   a= 3d = 3(2) = 6  B1 r= 12/8 = 1.5       A1
	Total 3mks
11	Use ratio theorem where m = 3 and n = −2 OM = np+mq             m+n        = −2( 1 2) 3 + 34 ( 5 ) −3  M1                          3+ −2       = ( −2 −4  ) −6 +12 ( 15 ) −9  = 10 ( 11 ) −15  M1                          1 Coordinate of M is (10, 11,−15)     A1
	Total 3mks
12	r² = (x − 9)²2 + (y − b)² CA = √((5−1)² + (5−3)²)      = √(4² + 2²)     = √20 units  (x−5)² + (y−5)² = (√20)²  x² − 10x + 25 + y² − 10y + 25 = 20  x² + y² − 10x − 10y + 50 = 20  x² + y² − 10x − 10y + 30 = 0  a² + 9² − 10a − 90 + 30 = 0  a² + 81 − 10a − 60 = 0  a² − 10a + 21 = 0  a = −b ± √(b² − 4ac)                   2a    = 10 ±√(100−(4×1×21))                   2   = 10 ± 4          2  ∴ a = 7 or 3
	Total 4mks
13	A can fill  1/3 of the tank in 1 hr. C can empty  1/5 of the tank in 1 hr. Fraction filled by both tanks in 1 hr =  1/3 − 1/5 =  2/15 Fraction filled in 4 hrs =  2/15 x 4 =  8/15 Empty fraction = 1 −  8/15 = 7/15   A1 Fraction filled in one hour by all pipes =  1/3 + 1/2 − 1/5 = 19/30  M1 Time to fill the empty fraction = 7/15 ÷  19/30 =  7/15 x 30/19 =  14/19hours  or=0.7368 hours  A1
	Total 4mks

20.  
    1. Complete the table below for y=sin 2x and y=sin ( 2x + 30) giving values to 2d.p.(2mks)

       x	0	15	30	45	60	75	90	105	120	135	150	165	180
       Sin 2x	0	0.5	0.87	1	0.87	0.5	0	-0.5	-0.87	-1	-0.87	-0.5	0
       Sin (2x+30)	0.5	0.87	1	0.87	0.5	0	-0.5	-0.87	-1	-0.87	-0.5	0	0.5

    2. Draw the graphs of y=sin 2x and y = sin (2x + 30) on the axis.     (4mks)
       
    3. Use the graph to solve Sin (2x + 30) − sin 2x = 0  (1mk)
       x = 37.5° or 125.5°
    4. Determine the transformation which maps  sin 2x onto sin (2x + 30)  (1mk) 
       Translation    
    5. State the period and amplitude of y = sin (2x + 30). (2mks)
       Period 180° amplitude 1unit
21. 
    
    
22. Let A = 62

    Marks	f	x	D=x-A	fd	d²	fd²
    45-49	3	47	-15	-45	225	675
    50-54	9	52	-10	-90	100	900
    55-59	13	57	-5	-65	25	325
    60-64	15	62	0	0	0	0
    65-69	5	67	25	25	25	125
    70-74	4	72	40	40	100	400
    75-79	1	77	15	15	225	225
    	F = 50			Efd - 120		=2650

    B1 d value
    B1 fd2 ratio
    1. Mean x = A +  Efd
                                Ef
          62 + -¹²⁰ / ₅₀ M1
       = 62 – 2.4  =  59.6  A1
    2. v = ² M1
         = 2560 − (120)² = 53 − 5.76   M1A1
               50       50
    3. s.d = 
           = √47.24
         = 6.873. A1
23.  
    1.  
       1. P(B) 
          = ³/₂₇ = ¹/₉ 
       2. P(P) or P(W)
          ⁹/₂₇ + ¹⁵/₂₇ = ²⁴/₂₇ = ⁸/₉
    2.  
       
       
       1. P(BB) = (³/₂₇ × ²/₂₆) + (³/₂₇ × ⁹/₂₆) + (³/₂₇ × ¹⁵/₂₆) + (⁹/₂₇ × ³/₂₆) + (¹⁵/₂₇ × ³/₂₆)
                   = ¹/₁₁₇ + ¹/₂₆ + ⁵/₇₈ + ¹/₂₆ + ⁵/₇₈ = ²⁵/₁₁₇ 
       2. P(BB) or P(PP) or P(WW)
          = (³/₂₇ × ²/₂₆) + (⁹/₂₇ × ⁸/₂₆) + (¹⁵/₂₇ × ¹⁴/₂₆)
          = ¹/₁₁₇ + ⁴/₃₉ + ³⁵/₁₁₇
          = ¹⁶/₃₉ 
24.  
    1.  
       1. R:T
          3:2
          Cost of 1kg of the blend A = (3 × 80) + (2 × 140)
                                                                   3 + 2
                                                    = 240 + 280
                                                              5
                                                    = Ksh. 104
          R → ³/₅ × 10 = 6kgs
          T → ²/₅ × 10 = 4kgs
          Blend X → R:T
                             1:2
          ∴    6     = ½
              4 + x
          4 + x = 12
                x = 8kgs
       2. Cost of 1kg of the blend A = (6 × 80) + (12 × 140)
                                                                     18
                                                     = Ksh. 120
          If 100% = Ksh. 120
          ∴ 120% = (120 × 120)
                                100
                       = Ksh. 144
    2. Cost of 1kg of the blend B = 104x + 140y
                                                            x + y
                                  ∴     130     =  104x + 140y
                                                             x + y
       130x + 130y = 104x + 140y
                      26x = 10y
                      13x = 5y
                      ^x/_y = ⁵/₁₃ 
       x : y
       5 : 13